Problem: $h(x)=-x^4-2x^3+72x^2-13$. On which intervals is the graph of $h$ concave up? Choose 1 answer: Choose 1 answer: (Choice A) A $\left(-\infty,-4\right)$ and $(3,\infty)$ (Choice B) B $\left(-\infty,-3\right)$ and $(4,\infty)$ (Choice C) C $(-3,\infty)$ only (Choice D) D $(-4,3)$ only
We can analyze the intervals where $h$ is concave up/down by looking for the intervals where its second derivative $h''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $h'$, we are analyzing $h''$. The second derivative of $h$ is $h''(x)=-12(x+4)(x-3)$. $h''(x)=0$ for $x=-4,3$. Since $h''$ is a polynomial, it's defined for all real numbers. Therefore, our points of interest are $x=-4$ and $x=3$. Our points of interest divide the number line into three intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $(-\infty, \llap{-}4)$ $( \llap{-}4,3)$ $(3,\infty)$ Let's evaluate $h''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h''(x)$ Verdict $\left(-\infty,-4\right)$ $x=-5$ $h''(-5)=-96<0$ $h$ is concave down $\cap$ $(-4,3)$ $x=0$ $h''(0)=144>0$ $h$ is concave up $\cup$ $(3,\infty)$ $x=4$ $h''(4)=-96<0$ $h$ is concave down $\cap$ In conclusion, the graph of $h$ is concave up over the interval $(-4,3)$ only.